Differential Geometry:Curves, Tangents, and Curvature

Differential Geometry of Curves and Surfaces, Do Carmo (Chapter 1)

Differentiable Curves

Definition:

A parameterized differentiable curve is a differentiable map α:I→Rn of an open interval I=(a,b) of the real line R into Rn.

Differentiable Curves

Examples:

• α(u)=(cos u , sin u , u)

• α(u)=(u , |u|)

• α(u)=(u3-4u , u2-4)

• α(u)=(u3 , u2)

( )uuuu ,sin,cos)( =α

( )4,4)( 23 −−= uuuuα ( )23,)( uuu =α( )uuu ,)( =α

Differentiable Curves

Note:

If φ(u) is a differentiable map from [a’,b’] to [a,b], then the function αoφ:[a’,b’]→Rn is also a parameterized differentiable curve tracing out the same points.

( )4,4)( 23 −−= uuuuα ( )23,)( uuu =α

Regular Curves

Given a parameterized curve, α(u), the tangentof α at a point u is the vector defined by the derivative of the parameterization, α’(u).

( )23,)( uuu =α

Regular Curves

Given a parameterized curve, α(u), the tangentof α at a point u is the vector defined by the derivative of the parameterization, α’(u).

The parameterized curve α(u) is regular if α’(u)≠0 for all u∈I.

( )23,)( uuu =α

Regular Curves

Given a parameterized curve, α(u), the tangentof α at a point u is the vector defined by the derivative of the parameterization, α’(u).

The parameterized curve α(u) is regular if α’(u)≠0 for all u∈I. This implies:

–There is a “well-defined” tangent lineat each point on the curve.

–The map α(u) is “injective”

( )4,4)( 23 −−= uuuuα

Regular Curves

We say that two parameterized, differentiable, regular, curves α:[a,b]→Rn and β:[a’,b’]→Rn

define the same differentiable curve if there exists a differentiable map φ:[a’,b’]→[a,b], with φ’(u)≠0, such that βoφ=α.

Regular Curves

We say that two parameterized, differentiable, regular, curves α:[a,b]→Rn and β:[a’,b’]→Rn

define the same differentiable curve if there exists a differentiable map φ:[a’,b’]→[a,b], with φ’(u)≠0, such that βoφ=α.We say that they define the same oriented differentiable curve if, in addition, φ’(u)>0.

Regular Curves

We say that two parameterized, differentiable, regular, curves α:[a,b]→Rn and β:[a’,b’]→Rn

define the same differentiable curve if there exists a differentiable map φ:[a’,b’]→[a,b], with φ’(u)≠0, such that βoφ=α.We say that they define the same oriented differentiable curve if, in addition, φ’(u)>0.From now on, we will talk about (oriented) curves as equivalence classes, and parameterized curves as instances of a class.

Regular Curves

Given a parameterized curve α(v), and given u∈I, the arc-length from the point u0 is:

∫=u

u

dvvus0

)(')( α

Regular Curves

Given a parameterized curve α(v), and given u∈I, the arc-length from the point u0 is:

If we partition the interval [u0,u] into N sub-intervals, setting ∆u=(u-u0)/N and ui=u0+i∆u, we can approximate the integral as:

∫=u

u

dvvus0

)(')( α

( )

( ) ( )

( ) ( )uuuu

uu

uu

uuus

N

iiiN

N

i

ii

N

N

iiN

∆∆

−=

∆∆−

=

∆=

∑

∑

∑

−

=+∞→

−

=

+

∞→

−

=∞→

1

01

1

0

1

1

0

lim

lim

'lim)(

αα

αα

α

α(u0)

α(ui) α(uN)

Regular Curves

Using the arc-length, we can integrate functions defined over the curve.Given a function f:Rn→R, the integral of f over the curve α can be obtained by integrating over the parameterization domain I:

( )∫∫ =∈

b

ap

duuufdppf )(')()( ααα

Regular Curves

Using the arc-length, we can integrate functions defined over the curve.Given a function f:Rn→R, the integral of f over the curve α can be obtained by integrating over the parameterization domain I:

Note: If the parameterizations α and β define the same curve, they define the same integral. So integrals are properties of curves.

( )∫∫ =∈

b

ap

duuufdppf )(')()( ααα

Regular Curves

Given a regular differentiable curve α, we can define the unit-tangent vector at a point α(u) as:

)(')(')(

ttut

αα

=

Regular Curves

Given a regular differentiable curve α, we can define the unit-tangent vector at a point α(u) as:

Note: If the parameterizations α and β define the same oriented curve, they define the same unit-tangent vectors. So unit-tangent vectors are properties of oriented curves.

)(')(')(

ttut

αα

=

Regular Curves

Given a regular differentiable curve α, we can define the unit-tangent vector at a point α(u) as:

Note: If the parameterizations α and β define the same oriented curve, they define the same unit-tangent vectors. So unit-tangent vectors are properties of oriented curves.Tangent lines are properties of (un-oriented) curves.

)(')(')(

ttut

αα

=

Curvature

The tangent line at a point α(u0) on the curve describes the best-fitting line to the curve.

α(u0)

Curvature

The tangent line at a point α(u0) on the curve describes the best-fitting line to the curve.We would like to extend this to finding the best-fit circle at a point.

α(u0)

Curvature

Q: How can we compute the (reciprocal of the) radius of a circle using only local information?

Q: How can we compute the (reciprocal of the) radius of a circle using only local information?A: Look at how quickly the tangent vectors are turning as we move between adjacent points.

Curvature

ϕ ϕp1

p0 p0

p1

t1

t0

t1

t0

Curvature

Q: How can we compute the (reciprocal of the) radius of a circle using only local information?A: Look at how quickly the tangent vectors are turning as we move between adjacent points.We use the fact that the length of the chord between p0 and p1 equals the distancebetween the tangents, multipliedby the circle’s radius: ϕ

p0

p1

t1

t0

rt1rt0

0101 ppttr −=−

Curvature

Thus, we can express the radius as:

01

011pptt

r −−

=

ϕp0

p1

t1

t0

rt1rt0

Curvature

Thus, we can express the radius as:

In the case of the curve, we can compute the curvature by looking at the limit as u1 goes to u0:

01

011pptt

r −−

=

)(')('

)()()()(

lim)(0

0

01

010

01 uut

uuutut

uuu ααα

κ =−−

=→

Curvature (2D)

For the case of an oriented curve in 2D, we can assign a sign to the curvature, making it positive if the center of the circle is to the left of the tangent and negative if it is to the right.

0<κ

0>κ

Curvature (2D)

For the case of an oriented curve in 2D, we can assign a sign to the curvature, making it positive if the center of the circle is to the left of the tangent and negative if it is to the right.

Note: If parameterizations α and β definethe same (oriented) curve, theydefine the same (signed)curvature. So curvature isa property of curves.

0<κ

0>κ

Curvature (2D)

Theorem:For a closed oriented curve α⊂R2, the integral of the curvature over α is an integer multiple of 2π:

∫∈

=α

πκp

ldpp 2)(

Curvature (2D)

Rough Proof:To prove this, we show that the integral of the curvature is equal to the integral of the change of angle traced out by the tangent indicatrix.

Curvature (2D)

Definition:Given an oriented differentiable curve α:I→R2, we can define the tangent indicatrix, a map from the interval I to the unit-circle defined by the tangent.

Curvature (2D)

Definition:Given an oriented differentiable curve α:I→R2, we can define the tangent indicatrix, a map from the interval I to the unit-circle defined by the tangent.

Curvature (2D)

Definition:Given an oriented differentiable curve α:I→R2, we can define the tangent indicatrix, a map from the interval I to the unit-circle defined by the tangent.

Curvature (2D)

Definition:Given an oriented differentiable curve α:I→R2, we can define the tangent indicatrix, a map from the interval I to the unit-circle defined by the tangent.

Curvature (2D)

Definition:Given an oriented differentiable curve α:I→R2, we can define the tangent indicatrix, a map from the interval I to the unit-circle defined by the tangent.

Curvature (2D)

Rough Proof:1. Using the tangent indicatrix, we can assign an

angle to each point on the curve, (the angle of the tangent vector).

Curvature (2D)

Rough Proof:2. With a bit of manipulation, we can show that

the change in the angle, as a function of arc-length, is equal to the curvature.

Curvature (2D)

Rough Proof:2. With a bit of manipulation, we can show that

the change in the angle, as a function of arc-length, is equal to the curvature.

Recall that the (signed) curvature is measured as the (signed) change in tangent length, divided by the change in arc-length.

)(')('

)(0

00 u

utu

ακ ±=

Curvature (2D)

Rough Proof:2. With a bit of manipulation, we can show that

the change in the angle, as a function of arc-length, is equal to the curvature.

Recall that the (signed) curvature is measured as the (signed) change in tangent length, divided by the change in arc-length.For small step sizes, the change in tangent length is equal to the change in angle:

ϕϕ ≈sin

Curvature (2D)

Rough Proof:3. Since the integral is the accumulation of

changes of angle, weighted by arc-length, and since we start and end with the same angle, the accumulated angle must be an integer multiple of 2π.

Curvature (2D)

Theorem:For a closed curve α⊂R2, the integral of the curvature over α is an integer multiple of 2π:

The integer l is the winding index of the curve.∫∈

=α

πκp

ldpp 2)(

Curvature (2D)

In defining the curvature, we implicitly used two different definitions:1. Curvature is the (signed) distance between

tangents divided by the distance between the corresponding points on the circle.

ϕp0

p1

t1

t0

01

010

01

lim)(pptt

ppp −

−=

→κ

Curvature (2D)

In defining the curvature, we implicitly used two different definitions:2. Curvature is the (signed) angle between

tangents divided by the arc-length between the corresponding points on the circle.

ϕp0=α(u0)

t1

t0

ϕ

∫→

=1

0

01

)('lim)( 0 u

u

uudvv

pα

ϕκp1=α(u1)

Curvature (2D)

In defining the curvature, we implicitly used two different definitions.In the limit, the two definitions are the same since the ratio of chord-length to arc-length goes to one as points get closer:

ϕ

)(22

sin2 3ϕϕϕ O+= sin(ϕ/2)ϕ/2

Curvature (2D)

In defining the curvature, we implicitly used two different definitions.In the limit, the two definitions are the same since the ratio of chord-length to arc-length goes to one as points get closer.But in the discretesetting, they givesrise to two differentnotions of curvature.

ϕ

sin(ϕ/2)ϕ/2

Discrete Curves

Definition:

A discrete curve is the set of points on the line segments between consecutive vertices in an ordered set of vertices {v0,…,vN}⊂Rn, such that for any index i, vi≠vi+1. vi

vi+1

Discrete Curves

The length of the discrete curve between vertexvi0 and vi is:

∑−

=+ −=

1

10

i

ijjji vvl

vivi+1

Discrete Curves

The length of the discrete curve between vertexvi0 and vi is:

Given a function assigning a value fi to edge vivi+1, we can compute theintegral of the function as:

∑−

=+ −=

1

10

i

ijjji vvl

vivi+1

∑−

=+ −=

1

01

N

iiii fvvF

Discrete Curves

The length of the discrete curve between vertexvi0 and vi is:

Similarly, given a function assigning a value fi to vertex vi, we can compute theintegral of the function as:

∑−

=+ −=

1

10

i

ijjji vvl

vivi+1

∑−

=

−+ −+−=

1

0

11

2

N

ii

iiii fvvvv

F

Discrete Curves

At any (point on an) edge vivi+1, we can define the tangent vector ti as the unit vector pointing from vi to vi+1:

ii

iii vv

vvt−−

=+

+

1

1

vivi+1

ti

ti+1ti-1

Discrete Curves

Q: Where and how should we define curvature?

vivi+1

ti

ti+1ti-1

Discrete Curves

Q: Where and how should we define curvature?

A: Since the tangent only changes at vertices we should define the curvature as a vertex value.

vivi+1

ti

ti+1ti-1

Discrete Curves

Q: Where and how should we define curvature?

A: Since the tangent only changes at vertices we should define the curvature as a vertex value.

We should define the value of the curvature as the change in the tangent aswe move through the vertex.

vivi+1

ti

ti+1ti-1

Discrete Curves

Note that we cannot define the tangent as the result of a limiting process:

vivi+1

[ ] [ ]

11

1

0

11

1

0

lim

)1()1(lim

−+

−

→∆

−+

−

→∆

−∆−

=

∆+∆−−∆+∆−−

=

ii

ii

t

iiii

ii

ti

vvttt

tvvttvvttt

κ

ti

ti+1ti-1

Discrete Curves

However, we can estimate it using the finite-differences using the edge centers (vi+vi+1)/2.

vivi+1

ti

ti+1ti-1

21 ii vv ++

Discrete Curves

However, we can estimate it using the finite-differences using the edge centers (vi+vi+1)/2.

1. Define the curvature in terms of change in tangents divided by the distance between edge centers: vi

vi+1

ti

ti+1ti-1

21 ii vv ++

[ ] [ ]2/)(2/)( 11

1

iiii

iii vvvv

tt+−+

−=

−+

−κ

ti

ti-1

21 ii vv ++

21−+ ii vv

Discrete Curves

However, we can estimate it using the finite-differences using the edge centers (vi+vi+1)/2.

2. Define the curvature in terms of change in tangents angle divided by the arc-length between edge centers: vi

vi+1

ti

ti+1ti-1

21 ii vv ++

2/)(2/)( 11

21

−+ −+−∠

=iiii

i vvvvttκ

ti

ti-1

21 ii vv ++

21−+ ii vv

Discrete Curves

In the limit, as we refine the tessellation of a differentiable curve, both methods define the same curvature.

Does it make a difference which one we use?

Discrete Curves

Does it make a difference which one we use?

Consider integrating the curvature over the closed curve:

∑−

=

−+ −+−=Κ

1

0

11

2

N

ii

iiii vvvvκ

vivi+1

ti

ti+1ti-1

Discrete Curves

Does it make a difference which one we use?

Consider integrating the curvature over the closed curve:

Using the second definition,we get:

∑−

=

−+ −+−=Κ

1

0

11

2

N

ii

iiii vvvvκ

∑−

=

∠=Κ1

021

N

itt

2/)(2/)( 11

21

−+ −+−∠

=iiii

i vvvvttκ

vivi+1

ti

ti+1ti-1

21 ii vv ++

Discrete Curves

Does it make a difference which one we use?

Consider integrating the curvature over the closed curve:

Using the second definition,we get the sum of angulardeficits over the vertices ofa closed polygon.

∑−

=

−+ −+−=Κ

1

0

11

2

N

ii

iiii vvvvκ

vivi+1

ti

ti+1ti-1

21 ii vv ++

Discrete Curves

Does it make a difference which one we use?

Consider integrating the curvature over the closed curve:

Using the second definition,we get the sum of angulardeficits over the vertices ofa closed polygon.

This has to be a multiple of 2π sothis definition conforms to the winding index.

∑−

=

−+ −+−=Κ

1

0

11

2

N

ii

iiii vvvvκ

vivi+1

ti

ti+1ti-1

21 ii vv ++

Discrete Curves

Does it make a difference which one we use?

Consider integrating the curvature over the closed curve:

Using the first definition wewould get a value for thecurvature integral that isnot a multiple of 2π, so thisdefinition does not conform tothe winding index.

∑−

=

−+ −+−=Κ

1

0

11

2

N

ii

iiii vvvvκ

vivi+1

ti

ti+1ti-1

21 ii vv ++